Lighting a LED with ambient RF (was candle)

a lossess loop would only reflect power, this would decrease the impedance of the power line nearby.

as someone else said, treat it as an air-core transformer

power line --------wwwwwwwwwwwwwwwwwwwww------------- --------wwwwwwwwwwwwwwwwwwwww------------- --------wwwwwwwwwwwwwwwwwwwww------------- ___nnnn__ | | | ___ | `-[___]-' load, (resistor, open, or short).

The hardest part of a solution is defining the problem. Your schematic leaves out the considerable loss to the ground underneath the fence.

I never did get along with dot and cross products. I prefer a more heuristic approach.

Given the frequency involved, a typical fence is WAY short of a wavelength. A more practical model might be a capacitive divider. The capacitance of the wire is tiny with respect to the total capacitance between the power line and the ground.

Open, the wire does mostly nothing. When loaded, the wire diverts some of the energy loss going into ground through the load before it gets to the ground. I claim that the net CHANGE in loss from the power line is miniscule... maybe some fraction (ratio of distances?) of the loss to the ground directly under the wire and way less than the total ground loss.

One experiment might be to measure the current between one end of the fence and ground. Open and short the other end. More when shorted >> inductive coupling. More when open >> capacitive coupling.

I propose that the capacitance dominates. Someone with math skills might have an opinion.

HV

wavelength.

The losses to ground (ground currents) aren't nearly as high as you think. The wire resistance is usually _the_ dominant loss factor.

Well i have seen the solvers operate and watched how the equations build up. I was at a high power seminar many years ago and magnetic induction dominated in every line studied.

?-)